Nucleophilic aromatic substitution (SNAr) template

Hydroxide adds to 1-chloro-2,4-dinitrobenzene at the carbon bearing Cl; the anion is delocalised onto the nitro groups, then chloride leaves to give 2,4-dinitrophenol.

Mechanism step by step

Addition–elimination through the Meisenheimer complex; nitro groups ortho and para hold the charge

1. Addition (slow) – OH⁻ attacks the C–Cl carbon; aromaticity is lost
2. Meisenheimer complex – The negative charge spreads onto a nitro oxygen
3. Elimination – Chloride leaves; aromaticity returns

Key points

• Needs strong electron-withdrawing groups ortho or para to the leaving group; a meta-NO₂ cannot take the charge.
• Addition is the slow step, so Ar–F reacts fastest (F ≫ Cl ≈ Br > I): Sanger’s reagent, 2,4-dinitrofluorobenzene, labels the N-terminus of peptides.
• Plain aryl halides do not do SN1 or SN2: the ring blocks backside attack and aryl cations are very unstable.
• In NaOH the phenol is present as its phenoxide; acid work-up gives 2,4-dinitrophenol.

Drawn with 112 atoms, 10 arrows, 7 shapes – every one editable in MolDraw.

Sources

• Clayden, Greeves & Warren, Organic Chemistry, 2nd ed., OUP 2012, ch. 22 Textbook
• Smith, March's Advanced Organic Chemistry, 8th ed., Wiley 2020, ch. 13 Textbook
• Meisenheimer, Justus Liebigs Ann. Chem. 1902, 323, 205 Original paper
• Bunnett & Zahler, Chem. Rev. 1951, 49, 273 Review